已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos²(x+θ/2)-√3,且0≤θ≤π,求使函数f(x)为偶函数的值
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![已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos²(x+θ/2)-√3,且0≤θ≤π,求使函数f(x)为偶函数的值](/uploads/image/z/5529666-66-6.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3D2sin%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89cos%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89%2B2%E2%88%9A3cos%26sup2%3B%EF%BC%88x%2B%CE%B8%2F2%EF%BC%89-%E2%88%9A3%2C%E4%B8%940%E2%89%A4%CE%B8%E2%89%A4%CF%80%2C%E6%B1%82%E4%BD%BF%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%E7%9A%84%E5%80%BC)
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos²(x+θ/2)-√3,且0≤θ≤π,求使函数f(x)为偶函数的值
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos²(x+θ/2)-√3,且0≤θ≤π,求使函数f(x)为偶函数的值
已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos²(x+θ/2)-√3,且0≤θ≤π,求使函数f(x)为偶函数的值
f(x)=sin(2x+θ)+2√3[1+cos(2x+θ)]/2-√3
=sin(2x+θ)+√3cos(2x+θ)
=2sin(2x+θ+π/3)是偶函数
f(-x)=f(x)
2sin(-2x+θ+π/3)=2sin(2x+θ+π/3)
所以-2x+θ+π/3=2kπ+2x+θ+π/3或-2x+θ+π/3=2kπ+π-(2x+θ+π/3)
-2x+θ+π/3=2kπ+2x+θ+π/3
4x=-2kπ
不是恒等式
-2x+θ+π/3=2kπ+π-(2x+θ+π/3)
2θ=2kπ+π/3
θ=kπ+π/6
所以θ=π/6